Solve the ODE with initial condition:

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Discussion Overview

The discussion revolves around solving a second-order ordinary differential equation (ODE) with specific initial conditions. Participants are exploring the correct application of methods to derive the solution and are addressing discrepancies between their results and an answer key.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant presents the ODE y'' - 10y' + 25 = 0 and the initial conditions y(0) = 0 and y'(1) = 12e^5, expressing confusion over their solution yielding y = (12/5)e^(5x) while the answer key states y = 2xe^(5x).
  • Another participant suggests that understanding the steps taken by the first participant is necessary to identify any errors.
  • A third participant shares their own solution process, stating they found the general solution to be y = c1e^(5x) + c2xe^(5x) and determined c1 = 0 from the initial condition y(0) = 0, leading to c2 = 12/5 from the condition y'(1) = 12e^5.
  • One participant points out the need to apply the product rule correctly when differentiating the second term of the general solution, indicating a potential source of error in the calculations.
  • A later reply acknowledges a mistake in the differentiation process and clarifies that the discussion is not about homework but rather exam preparation, while also expressing gratitude for the correction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct solution, as there are competing views regarding the application of differentiation rules and the resulting solutions. The discussion remains unresolved regarding the correct final answer.

Contextual Notes

Participants express uncertainty about the differentiation process and the implications of initial conditions on the constants in the general solution. There are indications of missing steps in the calculations that could affect the final outcome.

joker2014
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y''-10y'+25=0

Solve the ODE with initial condition:

y(0) = 0,

y' (1) = 12e^5 .

I keep getting y=12/5e^5x when c1=0 and c2=12/5 ... but Answer key says y=2xe^5x

what am I doing wrong?
 
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We have to see what you're doing to know what you're doing wrong.
 
axmls said:
We have to see what you're doing to know what you're doing wrong.
I solved it and got general solution of y=c1e^5x+c2xe5x and the derivative is y'=c1(5e^5x)+c2(5xe^5x)
for y(0)=0 i found that c1=0
in y'(1)=12e^5 i found that c2=12/5

which then gives final solution of y=(12/5)e^5x
 
You need to use the product rule on the second term: [tex](x e^{5x})' = x' e^{5x} + x (e^{5x})' = e^{5x} + 5x e^{5x}.[/tex]

Also, as a side note, it's advised that you use the homework section for any difficulties you're having in homework next time you have a question.
 
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axmls said:
You need to use the product rule on the second term: [tex](x e^{5x})' = x' e^{5x} + x (e^{5x})' = e^{5x} + 5x e^{5x}.[/tex]

Also, as a side note, it's advised that you use the homework section for any difficulties you're having in homework next time you have a question.
Ohmygodd! You are right I did a silly mistake!

This isn't homework, only studying for exam. But i thinkyes better to ask there.

Thank you
 

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